Question

How many 7 -digit numbers are there such that the digits are distinct integers taken from \((1,2, \ldots, 9)\) and the integers 5 and 6 do not appear together in either order?

Short answer

There are 151,200 valid 7-digit numbers.

Step-by-step solution

Understand the Problem

We need to find how many 7-digit numbers can be formed using the digits from 1 to 9, ensuring that each digit is distinct and the numbers 5 and 6 do not appear together.

Calculate Total Number of 7-Digit Combinations

We select 7 distinct digits from the 9 available (1 through 9). This can be done in \( \binom{9}{7} \) ways. For each selected set of 7 digits, we can arrange them in \( 7! \) ways. Therefore, the total number of combinations is \( \binom{9}{7} \times 7! \).

Calculate Combinations with Both 5 and 6

First, select the remaining 5 digits from the 7 available (excluding 5 and 6). This can be done in \( \binom{7}{5} \) ways. Consider 5 and 6 as one "block," which can be arranged in \( 2! \) ways within the 7-digit number. Thus, the total for this case is \( \binom{7}{5} \times 6! \times 2! \).

Subtract Cases with 5 and 6 from Total

Subtract the cases where both 5 and 6 are together from the total calculated in Step 2 to find the numbers where 5 and 6 do not appear together: \[(\binom{9}{7} \times 7!) - (\binom{7}{5} \times 6! \times 2!)\].

Calculate the Exact Numbers

Calculate each expression:- \( \binom{9}{7} \times 7! = 36 \times 5040 = 181440 \).- \( \binom{7}{5} \times 6! \times 2! = 21 \times 720 \times 2 = 30240 \).Thus, the numbers where 5 and 6 do not appear together is \( 181440 - 30240 = 151200 \).

Key concepts

Distinct Integers

When forming a number with distinct integers, each digit must be unique. This exercise requires the formation of 7-digit numbers using integers from 1 to 9. Therefore, **distinct integers** mean no repetition. For example, if you use the number "2", it can't appear again in any position within the same number.
Imagine trying to pick 7 different friends from a group of 9. Each friend corresponds to a digit. You cannot select the same friend twice, which means choosing 7 distinct individuals. This idea is central to combinatorics because it ensures diversity in choices. When selecting digits, it means we need to pull out 7 unique numbers from a possible 9, which leads us to next steps like permutation calculations.

Digit Selection

**Digit selection** is a crucial component in solving combinatorial problems. Here, we are selecting 7 digits from the total pool of 9 digits (from 1 to 9). To determine how many ways we can choose 7 digits, we use the binomial coefficient, written as \( \binom{9}{7} \).

• This notation represents how many combinations are possible when selecting 7 items from 9.
• The mathematical formula for computing such a binomial coefficient is \( \frac{9!}{7! \, 2!} \), resulting in 36 ways to choose 7 digits from 9.

Each combination of 7 digits will then be used in further permutations to create different potential numbers.

Permutations

Once 7 distinct digits have been selected, we must determine the number of ways to rearrange them to form different numbers. That's where **permutations** come in. Since the order of the digits matters in creating a different number, all unique arrangements of these 7 digits are considered.

• The formula for permutations of these 7 digits is \(7!\) (7 factorial), which means multiplying all whole numbers from 7 down to 1.
• Thus, \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\).

Total possible numbers using these selected digits are found by multiplying the combination count \(\binom{9}{7}\) by these permutations, giving us total possible configurations.

Combinatorial Restrictions

**Combinatorial restrictions** impose additional conditions on our selections. In this problem, although we are forming 7-digit numbers, specific conditions restrict us: the integers "5" and "6" cannot be used together.
To handle this restriction, we initially calculate the total number of 7-digit numbers without considering the restriction and then subtract cases where 5 and 6 appear together.

• Imagine grouping 5 and 6 as a single, inseparable 'block', changing the arrangement within the block for permutations (since 5 and 6 can switch places).
• With the block concept, you select 5 more integers from the remaining 7 (5 and 6 inclusive), and arrange all, maintaining the block as a unit.
• Calculate these exceptions using: \(\binom{7}{5} \times 6! \times 2! \), then subtract from the total.

By eliminating these restricted cases, we comply with the condition 5 and 6 are not together in the resulting numbers.